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3 years ago

What dissolved species are present in a solution of NaClO4?

1 answer:

Nadusha1986 [10]3 years ago

3 0

Assuming that the solution is simply an aqueous solution so that it is purely made of NaClO4 (the solute) and water (the solvent), then I believe the dissolved species would only be the ions of NaClO4, these are:

Na+

ClO4 -

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A phase is a part of a system with what type of composition?

TiliK225 [7]

Answer:

<u>uniform composition.</u>

Explanation:

A phase is a part of a system with uniform composition.

In physics and chemistry , a phase in a physically distinctive form of matter , such as solid , liquid , gas or plasma.
A phase matter is characterised by having relatively uniform chemical and physical properities . Phases are different from states of matter.
The states of matter (e.g ., solid , liquid , gas ) are phases but matter can exist in different phases and yet be in the same type of matter. For example , liquid mixtures can exist in multiple phases such as oily and aqueous phase.
some different types of phases are : solid, liquid,gas , plasma etc.
The term phase can also sometime mean describe equilibrium in phase diagram

6 0

3 years ago

HELPPPP STILLLL!!!! SCIENCE!!!!

Vadim26 [7]

Answer: the third one

Explanation:

4 0

3 years ago

The vapor pressure of water is 1.00 atm at 373 K, and the enthalpy of vaporization is 40.68 kJ mol!. Estimate the vapor pressure

Yuki888 [10]

Answer:

The vapor pressure at temperature 363 K is 0.6970 atm

The vapor pressure at 383 K is 1.410 atm

Explanation:

To calculate $\Delta H_{vap}$ of the reaction, we use clausius claypron equation, which is:

$\ln(\frac{P_2}{P_1})=\frac{\Delta H_{vap}}{R}[\frac{1}{T_1}-\frac{1}{T_2}]$

where,

$P_1$ = vapor pressure at temperature $T_1$

$P_2$ = vapor pressure at temperature $T_2$

$\Delta H_{vap}$ = Enthalpy of vaporization

R = Gas constant = 8.314 J/mol K

1) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =363 K

$T_2$ = final temperature =373 K

$P_2=1 atm, P_1=?$

Putting values in above equation, we get:

$\ln(\frac{1 atm}{P_1})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{363}-\frac{1}{373}]$

$P_1=0.69671 atm \approx 0.6970 atm$

The vapor pressure at temperature 363 K is 0.6970 atm

2) $\Delta H_{vap}=40.68 kJ/mol=40680 J/mol$

$T_1$ = initial temperature =373 K

$T_2$ = final temperature =383 K

$P_1=1 atm, P_2?$

Putting values in above equation, we get:

$\ln(\frac{P_2}{1 atm})=\frac{40680 J/mol}{8.314J/mol.K}[\frac{1}{373}-\frac{1}{383}]$

$P_2=1.4084 atm \approx 1.410 atm$

The vapor pressure at 383 K is 1.410 atm

8 0

3 years ago

Heating 235 g of water from 22.6°C to 94.4°C in a microwave oven requires 7.06 × 104 J of energy. If the microwave frequency is

Darya [45]

Answer: 3.69 × 10^27

Explanation:

Amount of energy required = 7.06 × 10^4 J

Frequency of microwave (f) = 2.88 × 10^10 s−1

Planck's constant (h) = 6.63 × 10^-34 Jᐧs/quantum

Recall ;

Energy of photon = hf

Therefore, energy of photon :

(6.63 × 10^-34)j.s× (2.88 × 10^10)s^-1

= 19.0944 × 10^(-34 + 10) = 19.0944×10^-24 J

Hence, number of quanta required :

(7.06 × 10^4)J / (19.0944 × 10^-24)J

= 0.369 × 10^(4 + 24) = 0.369×10^28

= 3.69 × 10^27

6 0

3 years ago

Determine the resulting pH when 12mL if 0.16M HCl are reacted with 32 mL if 0.24M KOH.

TEA [102]

Answer:

pH = 13.1

Explanation:

Hello there!

In this case, according to the given information, we can set up the following equation:

$HCl+KOH\rightarrow KCl+H_2O$

Thus, since there is 1:1 mole ratio of HCl to KOH, we can find the reacting moles as follows:

$n_{HCl}=0.012L*0.16mol/L=0.00192mol\\\\n_{KOH}=0.032L*0.24mol/L=0.00768mol$

Thus, since there are less moles of HCl, we calculate the remaining moles of KOH as follows:

$n_{KOH}=0.00768mol-0.00192mol=0.00576mol$

And the resulting concentration of KOH and OH ions as this is a strong base:

$[KOH]=[OH^-]=\frac{0.00576mol}{0.012L+0.032L}=0.131M$

And the resulting pH is:

$pH=14+log(0.131)\\\\pH=13.1$

Regards!

3 0

3 years ago